Electronics Engineering

Next, for NPN transistors connect the ohmmeter's positive lead to the base. Briefly touch the other meter lead to the transistor's Collector, then to the transistor's Emitter. Both connections should show a low-hm's reading: a diode-type connection. Then, connect the meter's negative lead to the transistor Base, and again touch the other meter lead to each of the other transistor pins. The meter should show open circuit or infinite ohms on both pins, indicating reverse-biased diodes.

If the transistor is PNP, just reverse the meter leads to perform the above tests. Touch the meter's negative lead to the Base pin, then verify that the Emitter and Collector pins behave as "turned on" diodes. Touch the meter's positive lead to the Base, and verify that the Collector and Emitter are acting like turned-off diodes.

First test for diode function in order to locate the transistor's Base pin. Connect your ohmmeter between the other two pins (no connection to Base, yet.) Wet your fingers and touch the Base at the same time as one of the other pins. Note the ohms reading if any. Then touch your fingers between Base and the other transistor pin. Was one ohms reading much lower than the other? If not, then reverse the ohmmeter leads. Then touch your fingers between Base and one pin, then Base and the other pin. When you find one ohmmeter reading which is much lower than all the others, then you've located the proper polarity for transistor amplification. During the lowest ohms reading, your wet fingers were connected between Base and Collector. The other transistor pin then has to be the Emitter. (You can also perform this whole test more reliably by using a 1K resistor in place of your wet fingers!)

First let's test for correct operation of a common N-channel depletion-mode MOSFET. Connect your ohmmeter between the FET's Source and Drain pins, with the positive meter lead on the Drain. While watching the ohms reading, touch the 9V battery's negative terminal to the Source or Drain, then use your wet fingers to briefly connect the battery's positive terminal to the transistor Gate pin. The ohms value between Source and Drain should become low. Remove the battery and briefly touch your wet fingers between Source, Drain, and Gate. The ohms reading should go high again. You've just applied and removed a charge to the floating Gate pin, switching the transistor on, then off again.

For testing a P-channel FET, just reverse all the polarities. Connect the ohmmeter to Source and Drain (positive meter lead to Source.) Use your wet fingers to touch the battery's negative terminal to the Gate pin (with battery positive touching Source or Drain.) That should make the ohms value go low.)

NOTE ON ALL OF THE ABOVE: I assume that the transistor is disconnected from any other circuit. If your transistor is soldered into a circuit board, most of these tests won't work.

the source voltage and the total impedance

A 'complex circuit' describes a category of circuit that is neither series, parallel, nor series-parallel. A relatively-simple example of a complex circuit is a Wheatstone Bridge. You cannot analyse or resolve a complex circuit using the techniques used to analyse and resolve series, parallel, or series-parallel circuit. Instead you must use one or other of the various electrical theorems. For example, to determine the currents flowing in a Wheatstone Bridge circuit, you could use Kirchhoff's Laws or Thevenin's Theorem.

For sine waves:

To calculate the RMS value of a sine wave, multiply the peak value by 0.707. The peak value is, of course, one half the peak-to-peak value. To go the other way, reverse the order of operations. That is, if you're starting with an RMS value, divide by 0.707 and then multiply by two to get the p-p value. Another way to convert from RMS to p-p is to multiply the RMS value by two square roots of two: RMS x 2 x SQR(2).

Or more simply, to convert from RMS to peak to peak voltage:

(RMS x 1.414) x 2=P-P

For example:

120vac x 1.414= 170vac

169.68vac x 2 = 339.36vac P-P

Where there is a significant reactive component in the characteristic of the cabling or load, we must also consider the effects of the X/R ratio. The real vs imaginary components of the impedance:

V(peak) = V(rms) * sqrt(2) * (1+ e(exp -pi/(X/R)))

where "exp" is the exponent to which the value of e is raised.

Reading the power of e above in English:

"e to the power of (minus pi divided by the X/R ratio)".

Take careful note of the top line of the first answer "For sine waves". More specifically the formula applies only to a signal which is a pure sine wave (a single frequency, no harmonics or other frequencies). As soon as you combine two or more sine wave signals of different frequencies, the ratio of peak voltage to RMS voltage depends strongly on the phases of the component signals. Too complicated to answer here, in full explanation.

Vrms = (Vp-p / 2 )/ sq.root (2) = Vp-p / 2.828 for sinusoidal waves..

*Note: If the signal or the waveform is not sinusoidal you'll have to derive from basic steps with a sound knowledge about integrations.

RMS = root mean square = square root (mean(square values of voltages))

A logic gate is a computer circuit with several inputs but only one output that can be activated by a particular combinations of inputs, a diagram that shows the major gates can be found here.

An example would be if you have 2 wires going to an AND gate and only 1 wire has a current (is on, 1) and the other one is 0, then your output would be 0 as you need wire 1 AND wire 2 to have a current, for the output to be 1, on.

A sampling gate, on the other hand is a circuit that produces an output only when first activated by a preliminary pulse. So if you have a current going through a wire and through a sampling gate, your output would be 0, unless you program the sampling gate to let the current through.

I'm sorry if this is a slightly plain answer, but if you would like more reading, then go to the Wikipedia.